We are given that $f$ is continuous on the interval $[0, 6]$, that the only solutions to $f(x) = 3$ are $x = 1$ and $x = 5$, and that $f(4) = 5$. We need to analyze the statements provided and determine which ones must be true.

Key facts:

1. Continuity of $f$: Since $f$ is continuous on $[0, 6]$, the Intermediate Value Theorem (IVT) can be applied.
2. $f(x) = 3$ has solutions only at $x = 1$ and $x = 5$. This means that for $x \in [0,6]$, $f(x) = 3$ only when $x = 1$ or $x = 5$.

Analysis of the statements:

Statement (i): $f(2) < 3$

• $f(1) = 3$ and $f(4) = 5$. Between $x = 1$ and $x = 4$, because $f$ is continuous and $f(4) = 5 > 3$, the function must increase at some point after $x = 1$.
• By the IVT, for $f(x) = 3$ to have no other solutions between $x = 1$ and $x = 5$, $f$ must remain strictly less than 3 between $x = 1$ and $x = 2$. Therefore, it is reasonable to conclude that $f(2) < 3$.

Statement (ii): $f(0) > 3$

• Since $f(1) = 3$ and there are no other points where $f(x) = 3$ on the interval $[0,1]$, $f$ must either be entirely above or entirely below 3 on this interval.
• If $f(0)$ were less than 3, then by continuity, it would cross 3 at some point before $x = 1$, which contradicts the fact that $x = 1$ is the only solution. Therefore, $f(0) > 3$.

Statement (iii): $f(6) < 3$

• Similarly, $f(5) = 3$, and there are no other points where $f(x) = 3$ on $[5, 6]$. Thus, $f(6)$ must be either greater than or less than 3.
• If $f(6)$ were greater than 3, $f$ would have to cross 3 again between $x = 5$ and $x = 6$, contradicting the fact that $x = 5$ is the only solution in this region. Therefore, $f(6) < 3$.

Conclusion:

All three statements (i), (ii), and (iii) must be true.

Thus, the correct answer is (H) all of them.

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Would you like any further clarification or details on the reasoning?

5 related questions:

1. What does the Intermediate Value Theorem state and how is it applied?
2. How does continuity influence the behavior of functions between specific points?
3. What would happen if there were additional points where $f(x) = 3$?
4. Why is the uniqueness of the solutions $f(x) = 3$ important in this problem?
5. How can we conclude that $f(0) > 3$ based on $f(1) = 3$?

Tip:

When analyzing continuous functions, the Intermediate Value Theorem is a powerful tool for deducing behavior between known points where the function takes specific values.